Answer:
Option (c) will be correct answer that is it will go 1.6 m
Explanation:
We have given that conveyor has the velocity u = 3.1 m/sec
Mass of the robot = 10 kg
static friction coefficient = 0.5 and kinetic friction coefficient = 0.3
Acceleration due to gravity g = 9.8 
Acceleration a = kinetic friction coefficient ×g = 0.3×9.8 = 2.94
Now according to third equation of motion
Finally velocity of the conveyor will be zero
So 
s = 1.6 m
So option (c) is correct option
Answer:
Explanation:
Given that,
Height of the bridge is 20m
Initial before he throws the rock
The height is hi = 20 m
Then, final height hitting the water
hf = 0 m
Initial speed the rock is throw
Vi = 15m/s
The final speed at which the rock hits the water
Vf = 24.8 m/s
Using conservation of energy given by the question hint
Ki + Ui = Kf + Uf
Where
Ki is initial kinetic energy
Ui is initial potential energy
Kf is final kinetic energy
Uf is final potential energy
Then,
Ki + Ui = Kf + Uf
Where
Ei = Ki + Ui
Where Ei is initial energy
Ei = ½mVi² + m•g•hi
Ei = ½m × 15² + m × 9.8 × 20
Ei = 112.5m + 196m
Ei = 308.5m J
Now,
Ef = Kf + Uf
Ef = ½mVf² + m•g•hf
Ef = ½m × 24.8² + m × 9.8 × 0
Ef = 307.52m + 0
Ef = 307.52m J
Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw
<span>At time t1 = 0 since the body is at rest, the body has an angular velocity, v1, of 0. At time t = X, the body has an angular velocity of 1.43rad/s2. Since Angular acceleration is just the difference in angular speed by time. We have 4.44 = v2 -v1/t2 -t1 where V and t are angular velocity and time. So we have 4.44 = 1.43 -0/X - 0. Hence X = 1.43/4.44 = 0.33s.</span>
Answer:
6.5 m/s^2
Explanation:
The net force acting on the yo-yo is
F_net = mg-T
ma=mg-T
now T= mg-ma
net torque acting on the yo-yo is
τ_net = Iα
I= moment of inertia (= 0.5 mr^2 )
α = angular acceleration
τ_net = 0.5mr^2(a/r)
Tr= 0.5mr^2(a/r)
(mg-ma)r=0.5mr^2(a/r)
a(1/2+1)=g
a= 2g/3
a= 2×9.8/3 = 6.5 m/s^2
Answer:
Explanation:
For the first ball, the moment of inertia and the kinetic energy is:
So, replacing, we get that:
At the same way, the moment of inertia and kinetic energy for second ball is:
So:
Then, 
is equal to 
, so:
Finally, solving for 
, we get:
